AP Calculus AB and BC

Limit of x/tan x as x Approaches 0

Q: Is x/tan x the same as x cot x?

Yes, since \cot x = 1/\tan x. So \lim_{x \to 0} x \cot x = 1 as well.

The limit of x over tangent x as x approaches zero is one. It is the reciprocal of the standard limit of tangent x over x, and since that limit is one and nonzero, the reciprocal is one as well.

\lim_{x \to 0} \frac{x}{\tan x} = 1

Settled by the reciprocal of the standard tangent limit.

Flipping a known limit

The standard result is $\lim_{x \to 0}\frac{\tan x}{x} = 1$ . Taking reciprocals is legitimate precisely because that limit is NOT zero: the reciprocal law needs a nonzero denominator limit.

\lim_{x \to 0}\frac{x}{\tan x} = \frac{1}{\lim_{x \to 0}\dfrac{\tan x}{x}} = \frac{1}{1} = 1

That caveat matters. Flipping $\lim_{x \to 0}\frac{\sin x}{x^{2}}$ , which is unbounded, would produce nonsense.

The direct route

Writing $\tan x = \frac{\sin x}{\cos x}$ gives $\frac{x}{\tan x} = \frac{x}{\sin x}\cos x$ . The first factor tends to 1 as the reciprocal of the standard sine limit, and $\cos x \to 1$ , so the product is 1.

Both routes are short. The reciprocal argument is faster to write; the direct one shows where the cosine goes.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

When is taking reciprocals of a limit allowed?

When the limit in the denominator exists and is nonzero. If it is zero, the reciprocal is unbounded and the law does not apply.

Is x/tan x the same as x cot x?

Yes, since $\cot x = 1/\tan x$ . So $\lim_{x \to 0} x \cot x = 1$ as well.